Croatian Mathematical Olympiad

Let $M(n)$ be the maximal possible number of good diagonals in a convex $n$-gon.
We will show that $M(n) = n - 2$ if $n$ is even and $M(n) = n - 3$ if $n$ is odd. For any $n$, we can draw all $n - 3$ diagonals from one vertex $A$ and one more diagonal joining two vertices adjacent to $A$ if $n > 3$. We show by mathematical induction that $M(n) \ge n - 2$ if $n$ is even. For $n = 4$ the claim is true as both diagonals in a convex quadrilateral are good. Let us assume that the claim is true for $n - 2$ and consider a convex $n$-gon $A_1A_2...A_n$. We draw diagonals $\overline{A_{n-2}A_n}$, $\overline{A_1A_{n-1}}$ and we use the assumption on $A_1A_2...A_{n-2}$, so $M(n) \ge M(n-2) + 2 \ge n - 4 + 2 = n - 2$.
We also prove by induction that $M(n) \le n - 2$ if $n$ is even and that $M(n) \le n - 3$ if $n$ is odd. Obviously, $M(3) = 0$ and $M(4) = 2$. Let us assume that the claim holds for all $n$ smaller than $k$.
Let us consider a choice of diagonals of a convex $k$-gon for which $M(k)$ is achieved.
First case: if there are two good diagonals that intersect, they divide the $k$-gon into 4 part, each having $a_i \ge 0$ vertices, not counting the endpoints of these diagonals (for $1 \le i \le 4$). Since there is no other segment that intersects these two diagonals, we have
$$M(k) = 2 + \sum_{i=1}^{4} M(a_i + 2) \le 2 + \sum_{i=1}^{4} (a_i + 2 - 2) = 2 + (k - 4) = k - 2.$$
If $k$ is odd, then at least one of the numbers $a_i$ must also be odd, so we get the bound $k - 3$ in this case.
Second case: if there are no two good diagonals that intersect, then there is automatically at most $k - 3$ good diagonals as that is the maximal number of diagonals that we can draw from one vertex.